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Note Mat. 35 (2015) no. 1, 1–13. doi:10.1285/i15900932v35n1p1

Postulation of general unions of lines and decorated lines

Edoardo Ballicoi

Department of Mathematics, University of Trento ballico@science.unitn.it

Abstract. A +line A ⊂ Pr, r ≥ 3, is the scheme A = L ∪ v with L a line and v a tangent vector of Pr supported by a point of L, but not tangent to L. Here we prove that a general disjoint union of lines and +lines has the expected Hilbert function.

Keywords:postulation, Hilbert function, lines, zero-dimensional schemes

MSC 2000 classification:primary 14N05, secondary 14H99

Introduction

Fix a line L ⊂ Pr, r ≥ 2, and P ∈ L. A tangent vector of Pr with P as its support is a zero-dimensional scheme Z ⊂ Pr such that deg(Z) = 2 and Zred= {P }. The tangent vector Z is uniquely determined by P and the line hZi spanned by Z. Conversely, for each line D ⊂ Pr with P ∈ D there is a unique tangent vector v with vred= P and hvi = D. A +line M ⊂ Pr supported by L and with a nilradical at P is the union v ∪ L of L and a tangent vector v with P as its support and spanning a line hvi 6= L. The set of all +lines of Prsupported by L and with a nilradical at P is an irreducible variety of dimension r − 1 (the complement of L in the (r − 1)-dimensional projective space of all lines of Pr containing P ). Hence the set of all +lines of Pr supported by L is parametrized by an irreducible variety of dimension r. Therefore the set of all +lines of Pr is parametrized by an irreducible variety of dimension 2(r − 1) + r = 3r − 1.

Now assume r ≥ 3. For all integers t ≥ 0 and c ≥ 0 let L(r, t, c) be the set of all disjoint unions X ⊂ Pr of t lines and c +lines. If (t, c) 6= (0, 0), then L(r, t, c) is an irreducible variety of dimension (t + c)(2r − 1) + cr. Fix any X ∈ L(r, t, c) and any integer k > 0. It is easy to check that h0(OX(k)) = (k + 1)(t + c) + c and hi(OX(k)) = 0 for all i > 0 (Lemma 2). A closed subscheme E ⊂ Pr is said to have maximal rank if for every integer k > 0 either h0(IE(k)) = 0 or h1(IE(k)) = 0, i.e. h0(IE(k)) = max{0, r+kr  − h0(OE(k))}.

iThis work is partially supported by MIUR and GNSAGA (INDAM) http://siba-ese.unisalento.it/ c 2015 Universit`a del Salento

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Theorem 1. Fix integers r ≥ 3, t ≥ 0 and c ≥ 0 such that (t, c) 6= (0, 0). If r ≥ 4, then assume that the characteristic is zero. Then a general X ∈ L(r, t, c) has maximal rank.

We prove Theorem 1 for r = 3 in arbitrary characteristic, while we assume characteristic zero if r ≥ 4. We also get intermediate results (e.g. Br,k) that may be useful as a sample of lemmas which may be proved with +lines. We see +lines as a tool to prove something involving the Hilbert function of unions of curves and fat points. For an alternative approach to such disjoint unions, see Remark 1.

1 Preliminaries

Remark 1. Fix a line L ⊂ Pn, n ≥ 2, and a linear system V ⊆ H0(OPn(k)).

Let L(1) be the first infinitesimal neighborhood of L in Pn, i.e. the closed sub- scheme of Pn with (IL)2 as its ideal sheaf. Let A be any +line with L as its support. For any closed subscheme B ⊂ Pn set V (−B) := {f ∈ V : f|B ≡ 0}.

The +line A gives independent conditions to V with the only restriction of that L is the support of L if either V (L) = {0} or dim(V (−A)) = dim(V (−L)) − 1.

A general +lines with L as its supports does not give independent conditions to V with the only restriction that L is its support if and only if V (−L) 6= {0} and V (−L(1)) = V (−L). Now assume dim(V (−L(1))) = dim(V (−L)) − γ for some γ > 0. The integer γ is the maximal number of tangent vectors v1, . . . , vγ of Pn supported by points of L and imposing independent conditions to V (−L) (with the restriction that their support is a point of L). So if we only need an integer t + c ≥ 2, t + c disjoint lines and x ≥ 2 tangent vectors supported by some of these lines we may decide to put more than one tangent vector on a single line.

Lemma 1. Let X ⊂ Pr be a closed subscheme such that the nilradical sheaf η ⊆ OX is supported by finitely many points. Set Y := Xred and fix k ∈ N.

Then:

(1) χ(OX(k)) = χ(OY(k)) + deg(η);

(2) h0(IX(k)) ≤ h0(IY(k)) ≤ h0(IX(k)) + deg(η);

(3) h1(IY(k)) ≤ h1(IX(k)) ≤ h1(IY(k)) + deg(η);

(4) h0(IX(k)) − h1(IX(k)) = h0(IY(k)) − h1(IY(k)) − deg(η).

Proof. By the definition of the reduction of a scheme the sheaf η is the ideal sheaf of Y in X. We have exact sequence (respectively of OX-sheaves and of OPr-sheaves):

0 → η → OX(k) → OY(k) → 0 (1)

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0 → IX(k) → IY(k) → η → 0 (2) Since η is supported by finitely many points, we have hi(η) = 0 for all i > 0 and deg(η) = h0(η). Use the cohomology exact sequences of (1) and (2). QED

Remark 2. Fix integers r ≥ 3, t ≥ 0 and c > 0. Fix A ∈ L(r, t, c), D ∈ L(r, t + c, 0) and set B := Ared.

(1) We have B ∈ L(r, t + c, 0). If A is general in L(r, t, c), then B is general in L(r, t + c, 0).

(2) Assume that D is general in L(r, t + c, 0) and fix a decomposition D = D1 t D2 with D1 ∈ L(r, t, 0) and D2 ∈ L(r, c, 0). Let E be a general element of L(r, 0, c) with Ered= D2. Then D1 is general in L(r, t, 0), D2

is general in L(r, c, 0) and D1∪ E is general in L(r, t, c).

Lemma 1 and Remark 2 give the following result.

Lemma 2. Fix integers r ≥ 3, t ≥ 0 and c > 0. Fix X ∈ L(r, t, c) and set Y := Xred. We have Y ∈ L(r, t + c, 0). If A is general in L(r, t, c), then B is general in L(r, t + c, 0). For each integer k > 0 we have h1(OX(k)) = 0, h0(OX(k)) = (t + c)(k + 1) + c, h0(IY(k)) − c ≤ h0(IX(k)) ≤ h0(IY(k)) and h1(IY(k)) ≤ h1(IX(k)) ≤ h1(IY(k)) + c.

For all integers r ≥ 3 and k ≥ 0 let Hr,k denote the following statement:

Assertion Hr,k, r ≥ 3, k ≥ 0: Fix (t, c) ∈ N2 \ {(0, 0)} and take a general X ∈ L(r, t, c). If (k + 1)t + (k + 2)c ≥ r+kk , then h0(IX(k)) = 0. If (k + 1)t + (k + 2)c ≤ r+kk , then h1(IX(k)) = 0.

Lemma 3. Fix a general X ∈ L(r, t, c). If 2t + 3c ≤ r + 1, then h1(IX(1)) = 0. If 2t + 3c ≥ r + 1, then h0(IX(1)) = 0.

Proof. Since the case c = 0 is obvious, we may assume c > 0 and use induction on c. Fix a general Y ∈ L(r, t, c− 1) and write X = Y t A with A a general +line of Pr. If 2t+3(c−1) ≥ r −1, we immediately see that h0(IY ∪Ared(1)) = 0. Hence h0(IX(1)) = 0. Hence we may assume 2t + 3(c − 1) ≤ r − 2. Let M ⊂ Pr be the (2t + 3c − 4)-dimensional linear subspace spanned by Y . Since A is general, it spans a plane N such that M ∩ N = ∅. Hence h0(IY ∪A(1)) = h0(IY(1)) − 3 =

r + 1 − 2t − 3c. QED

Remark 3. Fix an integer r ≥ 3. By the definition of maximal rank and the irreducibility of each L(r, t, c) Theorem 1 is true for the integer r if and only if all Hr,k are true. Since Hr,0 is obviously true, to prove Theorem 1 in Pr it is sufficient to prove Hr,k for all k > 0. Lemma 3 says that Hr,1 is true.

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Remark 4. Fix integers r ≥ 3 and k > 0 and suppose you want to prove Hr,k. Fix (t, c) ∈ N2\ {(0, 0)}. First assume t > 0 and (k + 1)(t + c) + c < r+kr .

Hence (k + 1)(t + c) + (c + 1) ≤ r+kr . Suppose that h1(IX(k)) = 0 for a general X ∈ L(r, t−1, c+1). Then h1(IY(k)) = 0 for a general Y ∈ L(r, t, c) (Lemma 1).

Now assume c > 0 and (k + 1)(t + c) + c > r+kr  and so (k + 1)(t + c) + (c − 1) ≥

r+k

r . Suppose that h0(IA(k)) = 0 for a general A ∈ L(r, t + 1, c − 1). Then h0(IB(k)) = 0 for a general B ∈ L(r, t, c). Therefore to prove Hr,k it is sufficient to test all (t, c) such that either (k + 1)(t + c) + c = k+rr 

or t = 0 and (k + 2)c < k+rr 

or c = 0 and (k + 1)t > r+kr . We do not need to test the pairs (t, 0) by [6]. Among the pairs (0, c) with (k + 2)c ≤ r+kr  it is sufficient to test the ones with r+kr  − k − 1 ≤ (k + 2)c ≤ r+kr .

For all integers r ≥ 3 and k ≥ 0 define the integers mr,k and nr,k by the relations

(k + 1)mr,k+ nr,k=r + k k



, 0 ≤ nr,k≤ k (3)

Remark 5. Fix integers r ≥ 3 and k > 0. Since mr,k ≤ k and k(k + 1) ≤

k+3

3  ≤ r+kr , we get mr,k≥ nr,k.

For all integers r ≥ 3 and k ≥ 0 set ur,k := d r+kr /(k + 2)e and vr,k :=

(k + 2)ur,k r+kr .

Notice that

(k + 2)(ur,k− vr,k) + (k + 1)vr,k=r + k r



(4) and that 0 ≤ vr,k≤ k + 1.

For all integers k > 0 let Ar,k denote the following assertion:

Assertion Ar,k, k > 0: Let X ⊂ Pr be a general union of vr,k lines and ur,k− vr,k +lines. Then h0(IX(k)) = 0.

2 The proof in P3

In this section we prove the case r = 3 of Theorem 1.

Lemma 4. Fix integers a ≥ 0, b ≥ 0, y ≥ 0. Let Z ⊂ Q be a general union of y tangent vectors. Then h0(IZ(a, b)) = max{0, (a + 1)(b + 1) − 2y} and h1(IZ(a, b)) = max{0, 2y − (a + 1)(b + 1)}.

Proof. By the semicontinuity theorem for cohomology ([5], III.12.8) it is suffi- cient to find a disjoint union W ⊂ Q of y tangent vectors such that h0(IW(a, b)) = max{0, (a + 1)(b + 1) − 2y}. It is obviously sufficient to do it for the integers

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y = b(a+1)(b+1)c and y = d(a+1)(b+1)/2e. First assume a odd. Let L0, . . . , Lb be b+1 distinct lines of type (0, 1). Let Ei⊂ Libe any disjoint union of (a+1)/2 tangent vectors. In this case we may take W = E1∪ · · · ∪ Eb. In the same way we conclude if b is odd. Hence we may assume that both a and b are even. If b = 0, then take y tangent vectors of L0. Similarly we conclude if a = 0. Hence we may assume a ≥ 2 and b ≥ 2 and use induction on a. It is obviously suffi- cient to check the integers y such that 2y ≥ (a + 1)(b + 1) − 1. Fix a smooth C ∈ |OQ(2, 2)|. C is a smooth elliptic curve and in particular it is irreducible.

Take a general S ⊂ C with ](S) = a+b. Let W ⊂ C be the union of the 2-points of C with the points of S as their support, i.e. the degree 2a+2b effective divisor of C in which each point of S appears with multiplicity two. Let W0 ⊂ Q be a union of y − a − b general tangent vectors. Set Z := W ∪ W0. By the inductive assumption we have h0(IW0(a − 2, b − 2)) = max{0, (a − 1)(b − 1) − 2y + 2a + 2b}, i.e. h0(IW0(a − 2, b − 2)) = max{0, (a + 1)(b + 1) − 2y} and h1(IW0(a − 2, b − 2)) = max{0, 2y − (a + 1)(b + 1)}. There are only finitely many (four in characteris- tic 6= 2, one or two in characteristic 2) line bundles R with R⊗2 = OC(a, b).

Since C has genus > 0 for general S the line bundle OC(S) is not one of them.

Hence W /∈ |OC(a, b)|. Since deg(W ) = deg(OC(a, b)), Riemann-Roch gives hi(C, OC(a, b)(−W )) = 0, i = 0, 1. Since ResC(Z) = W0, the Castelnuovo’s sequence gives hi(IZ(a, b)) = hi(IW0(a − 2, b − 2)). QED

We have u3,k := d(k + 3)(k + 1)/6e and v3,k := (k + 2)u3,k 3+k3 . Write k = 6m + b with 0 ≤ b ≤ 5. We have u3,6m = 6m2+ 4m + 1, v3,6m = 3m + 1, u3,6m+1= 6m2+ 6m + 2, v3,6m+1= 4m + 2, u3,6m+2= 6m2+ 8m + 3, v3,6m+2= 3m + 2, u3,6m+3 = 6m2+ 10m + 4, v3,6m+3 = 0, u3,6m+4 = 6m2 + 12m + 6, v3,6m+4 = m + 1, u3,6m+5 = 6m2 + 14m + 8, v3,6m+5 = 0. The construction below works (in particular Lemma 6) only because u3,6m+7−v3,6m+7≥ u3,6m+5 v3,6m+5 (both sides of the inequality are equal to u3,6m+5 = 6m2 + 14m + 8).

Without this inequality we would have needed a longer proof. In general we need u3,k+2− v3,k+2≥ u3,k− v3,k for all k > 0, but only in the case k = 6m + 5 the right hand side is not much bigger than the left hand side.

Let Q ⊂ P3 be a smooth quadric surface. We have Pic(Q) ∼= Z2 and we take two distinct, but intersecting, lines contained in Q as a basis of Pic(Q). We will call |OQ(1, 0)| and |OQ(0, 1)| the two rulings of Q and call any D ∈ |OQ(a, b)| a divisor (or a curve) of type (a, b). Fix a line L ⊂ Q, P ∈ L and let γ be the set of all +lines A ⊂ P3 with L as their support and P as the support of the nilradical sheaf of OA. The set γ is the complement of a point in a two-dimensional projective space (it is P(TPP3) \ P(TPL)). The set γ0 of all A ∈ γ contained in Q is the line P(TPQ) minus the point P(TPL). If A ∈ γ0, then A ⊂ Q and ResQ(A) = ∅. If A /∈ γ0, then A ∩ Q = L (as schemes) and ResQ(A) = {P } (as schemes). Now take A ⊂ Q, see L as a divisor of Q; we have ResL(A) = {P }.

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Lemma 5. Ar,1 and A3,2 are true.

Proof. Ar,1 is is true by Remark 3.

We have v3,2= 2 and u3,2− v3,2= 1. Take any B = L1t L2t L3 ∈ L(3, 3, 0).

B is contained in a unique quadric surface, Q0, and Q0 is a smooth quadric. Let A ⊂ P3be a general +line with L3as its support. We have L1∪L2∪A ∈ L(3, 2, 1).

Since A * Q0, we have h0(IL1∪L2∪A(2)) = 0. QED Of course, Ar,k makes sense only if ur,k− vr,k ≥ 0; this is the reason why we didn’t defined Ar,0. By the semicontinuity theorem for cohomology ([5], III.12.8) to prove A3,k it is sufficient to find A ∈ L(3, v3,k, u3,k − v3,k) such that h0(IA(k)) = 0. Fix an integer k > 0 and let X ⊂ P3 be a general union of v3,k lines and u3,k− v3,k +lines. We have h0(OX(k)) = (u3,k− v3,k)(k + 2) + v3,k(k + 1) = k+33 , the latter equality being true by the definition of the integer v3,k. Hence h1(IX(k)) = h0(IX(k)).

Lemma 6. A3,k ⇒ A3,k+2 for all k > 0.

Proof. We have 0 ≤ u3,k+2− u3,k ≤ k + 2.

(a) First assume v3,k+2 ≥ v3,k, i.e. k ≡ 0, 3, 4, 5 (mod 6). Notice that in all cases we have u3,k− v3,k ≤ u3,k+2− v3,k+2(we even have equality if k = 6m + 5, because u3,6m+7= 6m2+ 18m + 14, v3,6m+7= 4m + 6, u3,6m+5= 6m2+ 14m + 8 and v3,6m+5 = 0). Let L1 ⊂ Q be the union of v3,k+2− v3,k distinct lines of type (1, 0) and L2 ⊂ Q the union of (u3,k+2− v3,k+2) − (u3,k − v3,k) distinct lines of type (1, 0) with L1∩ L2 = ∅. Let A2 ⊂ Q be the union of (u3,k+2 v3,k+2)−(u3,k−v3,k) general +lines contained in Q and with the lines of L2as its support. Let S2⊂ L2be the support of the nilradical of OA2. Take a general Y ∈ L(3, v3,k, u3,k−v3,k). For general Y we have Y ∩(L1∪L2) = ∅ and so Y ∪A2∪L1 L(3, v3,k+2, u3,k+2− v3,k+2). By the semicontinuity theorem for cohomology ([5], III.12.8) it is sufficient to prove h0(IY ∪A2∪L1(k)) = 0. Since ResQ(Y ∪A2∪L1) = Y and h0(IY(k −2)) = 0, it is sufficient to prove h0(Q, IQ∩(Y ∪A2∪L1)(k +2)) = 0.

Since L1∪ L2 ⊂ (Y ∩ Q) ∪ A2 ∪ L1 and L1∪ L2 ∈ |OQ(u3,k+2− u3,k, 0)|, it is sufficient to prove h0(Q, IResL1∪L2((Y ∩Q)∪A2∪L1)(k +2−u3,k+2+u3,k, k +2)) = 0.

We have ResL1∪L2((Y ∩ Q) ∪ A2∪ L1) = (Y ∩ Q) ∪ S2. For general A2 the set S2 is a set containing a general point of (u3,k+2− v3,k+2) − (u3,k− v3,k) general lines of type (1, 0) and nothing else. Hence S2 may be considered as a general union of (u3,k+2 − v3,k+2) − (u3,k − v3,k) points of Q. For general Y the set Y ∩ Q is a general subset of Q with cardinality 2u3,k. Hence it is sufficient to check that ]((Y ∩ Q) ∪ S2) = h0(Q, OQ(k + 2 − u3,k+2+ u3,k, k + 2)), i.e.

2u3,k + (u3,k+2− v3,k+2) − (u3,k − v3,k) = (k + 3 − u3,k+2 + u3,k)(k + 3), i.e.

2u3,k+ (k + 2)(u3,k+2− u3,k) = (k + 3)2+ v3,k+2− v3,k. Taking the difference of

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(4) for the integer k0 = k + 2 from (4) and using that k+53  − k+33  = (k + 3)2 we get 2u3,k + (k + 2)(u3,k+2− u3,k) = (k + 3)2+ v3,k+2− v3,k, as wanted.

(b) Now assume v3,k+2 < v3,k, i.e. k ≡ 1, 2 (mod 6). Take a general Y ∈ L(3, v3,k, u3,k− v3,k) and write Y = E ∪ F with E ∈ L(3, v3,k+2, u3,k− v3,k) and F ∈ L(3, v3,k−v3,k+2, 0). For general Y we have h0(IY(k)) = 0 (by the inductive assumption) and Y ∩ Q is a general subset of Q with cardinality 2u3,k. For each line L ⊆ F fix one of the point PL∈ L∩Q and call vLa general tangent vector of Q at PL. Let AL= L ∪ vL⊂ P3be the +lines with L as its reduction, PLas the support of its nilradical and containing vL. Set G := ∪L∈FAL. Let M ⊂ Q be a union of u3,k+2− u3,k general lines of type (1, 0). Let N ⊂ Q be a general union of u3,k+2−u3,k+lines with M as the union of their support. We have E ∪G∪N ∈ L(3, v3,k+2, u3,k+2− v3,k+2). Since ResQ(E ∪ G ∪ N ) = Y and h0(IY(k)) = 0, it is sufficient to prove h0(Q, IQ∩(E∪G∪N )(k + 2, k + 2)). Since M ⊂ Q ∩ (E ∪ G ∪ N ), it is sufficient to prove h0(Q, IResM(Q∩(E∪G∪N ))(k +2−u3,k+2+u3,k, k +2)) = 0.

The scheme G ∩ Q is a general union δ of v3,k tangent vectors of Q and a general union of v3,k − v3,k+2 points of Q; we do not want to use here that general tangent vectors gives the maximal possible number of conditions to any linear system, because it requires characteristic zero ([4], [1], Lemma 1.4); however, since v3,k ≤ 2(k+2)/3, it is obvious that h1(Q, Iδ(k+2−u3,k+2+u3,k, k+2)) = 0;

alternatively, use Lemma 4. Set S := ResM(N ). The set S contains one point for each line of M and it is general with this condition. Since M is a general union of u3,k+2− u3,k lines of type (1, 0), S may be considered as a general subset of Q with its cardinality. The set E ∩ Q is a general subset of Q with cardinality 2u3,k− 2v3,k+2. Since 2(v3,k− v3,k+2) + (v3,k− v3,k−2) + (u3,k+2− u3,k) + 2(u3,k v3,k+2) = (k + 3 − u3,k+2+ u3,k)(k + 3), we are done. QED

Lemma 7. For all integers k > 0 and c > 0 such that c(k + 2) ≤ k+33  we have h1(IX(k)) = 0 for a general X ∈ L(3, 0, c).

Proof. If k = 1, then c = 1. The lemma is obvious in this case.

Now assume k = 2. It is sufficient to do the case c = 2. Take A = A1∪ A2 L(3, 0, 2) with L1 and L2 two different lines of type (1, 0) of Q, A1 ⊂ Q and general with these restrictions, A2 * Q and general among the +lines supported by Q. We get h0(Q, OQ∩A(2)) = 2 and h0(IResQ(A)) = 0, because ResQ(A) 6= ∅.

From now on we assume k ≥ 3. Lemmas 5 and 6 give that A3,k−2 and A3,k are true. We have c ≤ u3,k and c ≤ u3,k − 1 if v3,k > 0. If c ≤ u3,k − v3,k, then we may use A3,k. In particular we are done if v3,k = 0. Hence we may assume v3,k > 0. In this case it is sufficient to do the case c = u3,k − 1. Fix a general Y ∈ L(3, v3,k−2, u3,k−2− v3,k−2). We have hi(IY(k − 2)) = 0, i = 0, 1, by A3,k−2. We mimic part (b) of the proof of Lemma 6. Write Y = E t F with E ∈ L(3, 0, u3,k − v3,k) and F ∈ L(3, v3,k, 0). For each line L ⊆ F fix

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one of the point PL ∈ L ∩ Q and call vL a general tangent vector of Q at PL. Let AL = L ∪ vL ⊂ P3 be the +line with L as its reduction, PL as the support of its nilradical and containing vL. Set G := ∪L∈FAL. Let M ⊂ Q be a union of u3,k+2− u3,k− 1 general lines of type (1, 0). Let N ⊂ Q be a general union of u3,k+2 − u3,k − 1 +lines with M as the union of their support. Take X := E ∪ G ∪ N ∈ L(3, 0, u3,k− 1). Since v3,k ≤ k + 1, as in part (b) of the proof of Lemma 6 we get h1(Q, IX∩Q(k)) = 0 and hence h1(IX(k)) = 0. QED

Proof of Theorem 1 for r = 3: It is sufficient to prove H3,k for all k ≥ 2 (Remark 3). Fix an integer k > 0. It is sufficient to check the Hilbert function in degree k of a general element of L(3, t, c) with either t = 0 and (k + 2)c ≤ k+33  or (k + 1)t + (k + 2)c = k+33 

(Remark 4). By Lemma 7 it is sufficient to check the pairs (t, c) with t > 0 and (k + 1)t + (k + 2)c = k+33 , i.e. (since the integers k + 1 and k + 2 are coprime) the pairs (t, c) with t = v3,k + (k + 2)α, c = u3,k− v3,k− (k + 1)α for some non-negative integer α such that (k + 1)α ≤ u3,k− v3,k. By [6] we may assume c > 0. By Lemma 6 we may assume t > v3,k. Since v3,2= 2, u3,2= 3 and A3,2 is true (Lemma 5), we may assume k ≥ 3. By induction on k we may assume that hi(IW(k − 2)) = 0, i = 0, 1, for a general W ∈ L(3, t0, c0) for all non-negative integers t0, c0such that (k−1)t0+kc0 = k+13 .

Fix a general Y ∈ L(3, m3,k−2− n3,k−2, n3,k−2). We have hi(IY(k − 2)) = 0, i = 0, 1.

Claim 1: We have t + c ≥ m3,k−2.

Proof of Claim 1: Assume t + c ≤ m3,k−2− 1, i.e. assume (k − 1)(t + c) + k − 1 ≤ k+13 . Since (k + 1)t + (k + 2)c = k+33 , we get (k + 2)/(k − 1) >

k+3

3 / k+13  = (k + 3)(k + 2)/k(k − 1), a contradiction.

Claim 2: We have c ≥ n3,k−2.

Proof of Claim 2: If k − 2 ≡ 0, 1 (mod 3), then n3,k−2 = 0. If k − 2 ≡ 2 (mod 3), then n3,k−2= (k − 1)/3. Hence we may assume k ≡ 4 (mod 3). Since n3,k = 0, c > 0 and (k + 1)t + (k + 2)c = k+33 , we get c = β(k + 1) for some integer β > 0. Hence c ≥ k + 1 > n3,k−2.

Notice that m3,k−2 ≥ n3,k−2. Fix a general Y ∈ L(3, m3,k−2−n3,k−2, n3,k−2).

By the inductive assumption we have hi(IY(k − 2)) = 0, i = 0, 1. Set e :=

t + c − m3,k−2. Claim 1 gives e ≥ 0. Take a general union M ⊂ Q of e lines of type (1, 0).

(a) Assume c − n3,k−2 ≤ e. Claim 2 gives c − n3,k−2 ≥ 0. Write M = M1t M2 with M2 a union of c − n3,k−2 lines and M1 a union of e − (c − n3,k−2) lines. Let A2 ⊂ Q be a general union of c − n3,k−2 +lines with the lines in M2 as their support. Let S2 be the support of the nilpotent sheaf of A2. Since A2 is general, S2 is obtained taking for each line L ⊆ M2 a general point of L. Set X := Y ∪ M1∪ A2. Since X ∈ L(3, t, c), it is sufficient to prove that

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h1(IX(k)) = 0. Since ResQ(X) = Y and h1(IY(k − 2)) = 0, it is sufficient to prove that h1(Q, IX∩Q(k)) = 0. Since ResM(X ∩Q) = (Y ∩Q)∪S2, it is sufficient to prove that h1(Q, I(Y ∩Q)∪S2(k − e, k)) = 0. Since M2 is general and for each line L ⊂ M2the set S2∩L is a general point of L, S2is a general subset of Q with cardinality c − n3,k−2. Since Y is general, the set (Y ∩ Q) ∪ S2 is a general subset of Q of cardinality 2m3,k−2+ c − n3,k−2. Since (k − 1)m3,k−2+ n3,k−2 = k+13 , (k + 1)t + (k + 2)c = k+33 , k+33  − k+13  = (k + 1)2 and e = t + c − m3,k−2, we have 2m3,k−2+ (k + 1)e + c − n3,k−2 = (k + 1)2, i.e. ](S2 ∪ (Y ∩ Q)) = (k + 1)(k + 1 − e) = h0(Q, OQ(k − e, k)). Hence hi(Q, I(Y ∩Q)∪S2(k − e, k)) = 0, i = 0, 1.

(b) Now assume c − n3,k−2 > e. For each line R ⊆ M fix a general OR ∈ R and call vR a general tangent vector of Q with OR as its support.

Set R+ := R ∪ vR, M+ := ∪R⊆MR+ and S := ∪R⊆MOR. Since M is general and each OR is general in R, S is a general subset of Q with cardinality e.

We have M+ ⊂ Q and hence M+∩ Q = M+ and ResQ(M+) = ∅. Set g :=

c − n3,k−2− e. Since e = t + c − m3,k−2, we get m3,k−2− n3,k−2= g + t > t. Write Y = Y1 t Y2 with Y2 ∈ L(3, 0, n3,k−2) and Y1 ∈ L(3, m3,k−2− n3,k−2, 0). Since m3,k−2− n3,k−2= g + t ≥ t, we may write Y1 = Y3t Y4 with Y3 ∈ L(3, t, 0) and Y4 ∈ L(3, g, 0). For each line L ⊆ Y4 fix one of the two points, say OL, of L ∩ Q and let wL be a general tangent vector of Q with OL as its support; set L+:=

L ∪ wL∈ L(3, 0, 1). Set Y4+:= ∪L⊆Y4L+ and X0 := M+∪ Y4+∪ Y2∪ Y3. Since X0 ∈ L(3, t, c), it is sufficient to prove that h1(IX0(k)) = 0. Since ResQ(X0) = Y and h1(IY(k−2)) = 0, by the Castelnuovo’s sequence it is sufficient to prove that h1(Q, IX0∩Q(k)) = 0. The scheme X0∩ Q is the union of M+and (Y3∪ Y4+) ∩ Q.

We have ResM(M+∪ (Y3∪ Y4+) ∩ Q) = S ∪ ((Y \ Y4) ∩ Q) ∪S

L⊆Y4wL. Since Y ∩ Q is a general subset of Q with cardinality 2m3,k−2, the scheme Z :=

S ∪ ((Y \ Y4) ∩ Q) ∪S

L⊆Y4wL is a general union of g tangent vectors of Q and e + 2m3,k−2− g points of Q. By the Castelnuovo’s sequence it is sufficient to prove that h1(Q, IZ(k − e, k)) = 0. By Lemma 4 it is sufficient to check that deg(Z) ≤ (k − e + 1)(k + 1). By (3) for the integers r = 3 and k0= k − 2 and the equality (k + 1)t + (k + 2)c = k+33  we get deg(Z) = (k + 1)(k − e + 1). QED

3 When r > 3

In this section we prove Theorem 1 for all integers r ≥ 4. For numerical reasons this is easier than in the case r = 3 (as it was in [6] and [2]). The proof in characteristic zero is very short and we will only give it.

Lemma 8. For all integers r ≥ 3 and k ≥ 2 we have mr,k−1< ur,k.

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Proof. We have kmr,k−1 k+r−1r  and (k + 2)ur,k r+kr . Note that

r + k r

.r + k − 1 r



= (r + k)/k

and that (r + k)k > (k + 2)k for all r ≥ 3. QED We need the following assumption Br,k:

Br,k, r ≥ 4, k > 0. Fix a hyperplane H ⊂ Pr. There is X ∈ L(r, mr,k nr,k, nr,k) such that the support of the nilradical sheaf of X is contained in H and h0(IX(k)) = 0.

For all X ∈ L(r, mr,k − nr,k, nr,k) we have h0(OX(k)) = r+kr 

and so h1(IX(k)) = h0(IX(k)).

Lemma 9. For all integers r ≥ 4 and k ≥ 2 we have mr,k≥ mr,k−1. Proof. We have

mr,k−1+ (k + 1)(mr,k− mr,k−1) + nr,k− nr,k−1 =r + k − 1 r − 1



(5) Assume mr,k≤ mr,k−1− 1. Since nr,k− nr,k−1≤ k, (6) gives

mr,k−1− 1 ≥r + k − 1 r − 1



Since kmr,k−1 r+k−1r  and k r+k−1r−1  = r r+k−1r , we get −k ≥ (r − 1) r+k−1r−1 ,

Lemma 10. Fix an integer r ≥ 4 and assume that Theorem 1 is true in Pr−1. Then Br,k is true for all k > 0.

Proof. Br,1is true by Remark 3. Hence we may assume k ≥ 2 and that Br,k−1is true. Fix Y ∈ L(r, mr,k−1−nr,k−1, nr,k−1) such that the support of the nilradical sheaf of Y is contained in H and h0(IY(k − 1)) = 0. By the semicontinuity theorem for cohomology ([5], III.12.8) we may assume that Y is general among the elements of L(r, mr,k−1− nr,k−1, nr,k−1) whose nilradical sheaf is supported by points of H. Hence we may assume that no irreducible component of Yred is contained in H, that Yred is a general subset of H with cardinality mr,k−1

and that for each +line A ⊂ Y , say A = L ∪ vL, the tangent vector vL of A is not contained in H. The latter assumption implies Yred∩ H = Y ∩ H (scheme- theoretic intersection) and ResH(Y ) = Y . We have mr,k ≥ mr,k−1 (Lemma 9).

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(a) In this step we assume nr,k < nr,k−1. Let F ⊂ H be a general union of mr,k − mr,k−1 lines, with the only restriction that exactly nr,k−1− nr,k of them contain a point of Yred∩ H. We have mr,k−1− nr,k−1− (nr,k−1− nr,k) + mr,k− mr,k−1 = mr,k− 2nr,k−1 + nr,k. The scheme Y ∪ F is a disjoint union of nr,k−1− nr,k sundials, nr,k +lines and mr,k − 2nr,k−1 + nr,k lines. Since a sundial is a flat limit of a family of elements of L(r, 2, 0) ([2]), it is sufficient to prove h0(IY ∪F(k)) = 0. Since the set Yred∩ H is general in H, F may be considered as a general union of lines. Hence F has maximal rank. By (5) we have h1(H, IF(k)) = 0 and h0(H, IF(k)) = mr,k−1 − nr,k−1 + nr,k. Since for fixed Yred∩ H ∩ F we may deform the other components of Y so that the other mr,k−1 − nr,k−1 + nr,k points of Yred∩ (H \ F ) are general in H, then hi(H, IH∩(Y ∪F )(k)) = 0. Castelnuovo’s sequence gives h0(IY ∪F(k)) = 0.

(b) In this step we assume nr,k ≥ nr,k−1 and mr,k− nr,k ≥ mr,k−1 nr,k−1. Let E ⊂ H be a general union of mr,k− nr,k− (mr,k−1− nr,k−1) lines and nr,k − nr,k−1 +lines. We have Y ∪ E ∈ L(r, mr,k − nr,k, nr,k) and the support of the nilradical sheaf of Y ∪ E is contained in H. By (5) we have h0(E, OE(k)) + deg(Y ∩ H) = r+k−1r−1 . Since Theorem 1 is true in Pr−1, we have h1(H.IE(k)) = 0. Since Y ∩ H is a general union of mr,k−1 points of H, (5) implies hi(H, I(Y ∪E)∩H(k)) = 0.

(c) In this step we assume nr,k≥ nr,k−1and mr,k− nr,k< mr,k−1− nr,k−1. Therefore g := nr,k−nr,k−1−(mr,k−mr,k−1) > 0. Since nr,k≤ k, we have g ≤ k.

Since kmr,k−1+nr,k−1 = r+k−1r  and nr,k−1≤ k−1, we have g ≤ mr,k−1−nr,k−1. Take a general union G ⊂ H of mr,k−mr,k−1+lines. Since Theorem 1 is assumed to be true in Pr−1, G has maximal rank. By (5) we have h1(H, IG(k)) = 0 and h0(H, IG(k)) = mr,k−1+ g. Write Y = Y1t Y2t Y3 with Y3 ∈ L(r, 0, nr,k−1), Y1 ∈ L(r, mr,k−1−nr,k−1, 0) and Y2 ∈ L(r, g, 0). For each line L ⊆ Y2let vLbe the general tangent vector of H with L ∩ H as its support. Set A2 := ∪L⊆Y2(L ∪ vL).

Since U := Y1∪ A2∪ Y3∪ G ∈ L(r, mr,k− nr,k, nr,k), it is sufficient to prove that h0(IU(k)) = 0. We have ResH(U ) = Y , because each vLis contained in H. The scheme U ∩ H is the union of G, mr,k−1− nr,k−1− g general points of H and g general tangent vectors of H. Hence hi(H, IU ∩H(k)) = 0, i = 0, 1 ([1], Lemma

1.4). QED

Proof of Theorem 1 for r > 3: Let H ⊂ Pr be a hyperplane. We use induction on r, the starting case being the one with r = 3 proved in section 2.

Hence we assume Theorem 1 in H ∼= Pr−1 for all L(r − 1, t0, c0). By Remark 3 it is sufficient to prove Hr,k for all k > 0. Hr,1 is true (Remark 3). Hence we may assume k ≥ 2 and that Hr,k−1 is true. By Remark 4 it is sufficient to prove Hr,k for the pairs (t, c) such that either t = 0 and r+kr  − k − 1 ≤ c(k + 2) ≤ r+kr  or t(k + 1) + (k + 2)c = r+kr 

and c > 0. If r+kr  − k − 1 ≤ c(k + 2) ≤

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